Question

Solve the following pairs of equations:

$\frac{x}{a}+\frac{y}{b}=a+b;\frac{x}{a^2}+\frac{y}{b^2}=2;a,b\ne 0$

• A. $x=2a^2,y=2b^2$
• B. $x=a^3,y=2b^2$
• C. $x=a^3,y=3b^2$
• D. $x=a^2,y=b^2$

Answer 1

Answer: (D)

$x=a^2,y=b^2$

Solution:-

$\frac{x}{a}+\frac{y}{b}=a+b⟶eq.\left(1\right)$

$\frac{x}{a^2}+\frac{y}{b^2}=2⟶eq.\left(2\right)$

$\text{By multiplying eq.}\left(1\right)by\frac{1}{a},\text{we get}$

$\frac{x}{a^2}+\frac{y}{ab}=1+\frac{b}{a}⟶eq.\left(3\right)$

On subtracting eq. $\left(3\right)from\left(2\right)$, we get

$\frac{x}{a^2}+\frac{y}{b^2}-\frac{x}{a^2}-\frac{y}{ab}=2-1-\frac{b}{a}$

$\Rightarrow \frac{y}{b}(\frac{1}{b}-\frac{1}{a})=1-\frac{b}{a}\Rightarrow \frac{y}{b}=b$

$\Rightarrow y=b^2$

On putting the value of y in eq. $\left(1\right)$, we get

$\frac{x}{a}+\frac{b^2}{b}=a+b\Rightarrow x=a^2$

Hence, $x=a^2\&y=b^2$.